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Taya2010 [7]
3 years ago
13

Say a hacker has a list of n distinct password candidates, only one of which will successfully log her into a secure system. a.

If she tries passwords from the list at random, deleting those passwords that do not work, what is the probability that her first successful login will be (exactly) on her k-th try?
Mathematics
1 answer:
alexdok [17]3 years ago
7 0

Answer:

The probability is \frac{1}{n}

Step-by-step explanation:

If she has n distinct password candidates and only one of which will successfully log her into a secure system, the probability that her first first successful login will be on her k-th try is:

If k=1

P = \frac{1}{n}

Because, in her first try she has n possibles options and just one give her a successful login.

If k=2

P=\frac{n-1}{n} *\frac{1}{n-1} =\frac{1}{n}

Because, in her first try she has n possibles options and n-1 that are not correct, then, she has n-1 possibles options and 1 of that give her a successful login.

If k=3

P=\frac{n-1}{n} *\frac{n-2}{n-1} *\frac{1}{n-2} = \frac{1}{n}

Because, in her first try she has n possibles options and n-1 that are not correct, then, she has n-1 possibles options and n-2 that are not correct and after that, she has n-2 possibles options and 1 give her a successful login.

Finally, no matter what is the value of k, the probability that her first successful login will be (exactly) on her k-th try is 1/n

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The monthly expenses are found below:


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Adding all the expenses, will give us a total of $1695.
 
To get the percent which is allocated for movie and the baby sitting:


Add the amount for the two then divide that to the total then multiply it by 100%
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A The number of tropical fish that an aquarium can hold depends on the volume of the fish tank. The interior dimensions of a fis
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Step-by-step explanation:

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x is a number between 8 and 12 exclusive.​

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The value of x.

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I'm going to separate it into parts

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(y - 4)(y^2 - 5y + 6)

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